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(A) $1$. Substance $C$ decolorizes Baeyer's reagent,which indicates that $C$ is an alkene.$2$. Ozonolysis of $C$ gives $HCHO$ (formaldehyde),which mea

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$C_2H_6$

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(A) $1$. Substance $C$ decolorizes Baeyer's reagent,which indicates that $C$ is an alkene.$2$. Ozonolysis of $C$ gives $HCHO$ (formaldehyde),which means $C$ must contain a terminal methylene group $(CH_2=)$.$3$. Since $C$ is formed from $B$ by dehydrohalogenation (using alcoholic $KOH$),$B$ must be an alkyl halide.$4$. If $C$ is ethene $(CH_2=CH_2)$,then $B$ is ethyl chloride $(CH_3CH_2Cl)$,and $A$ is ethane $(C_2H_6)$.$5$. Reaction sequence: $C_2H_6$ $\xrightarrow{Cl_2/h
u} C_2H_5Cl (B)$ $\xrightarrow{alc. KOH} CH_2=CH_2 (C)$ $\xrightarrow{O_3/Zn, H_2O} 2HCHO$.

Column-$I$	Column-$II$
$A$. $S_N1$ reaction	$1$. vic-dibromides
$B$. Chemicals in fire extinguisher	$2$. gem-dihalides
$C$. Bromination of alkenes	$3$. Racemisation
$D$. Alkylidene halides	$4$. Saytzeff rule
$E$. Elimination of $HX$ from alkyl halide	$5$. Chlorobromocarbons

Chlorination of substance $A$ gives substance $B$. Substance $B$ reacts with alcoholic $KOH$ to give substance $C$,which decolorizes Baeyer's reagent. Ozonolysis of substance $C$ gives $HCHO$. What is substance $A$?

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Match the reactions given in Column-$I$ with the names given in Column-$II$.

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